$a,PTPƯ:Mg+2HCl\xrightarrow{} MgCl_2+H_2↑$
$n_{Mg}=\dfrac{7,2}{24}=0,3mol.$
$Theo$ $pt:$ $n_{H_2}=n_{Mg}=0,3mol.$
$⇒V_{H_2}=0,3.22,4=6,72l.$
$b,Theo$ $pt:$ $n_{HCl}=2n_{Mg}=0,6mol.$
$⇒m_{ddHCl}=\dfrac{0,6.36,5}{20\%}=109,5g.$
$c,Theo$ $pt:$ $n_{MgCl_2}=n_{Mg}=0,3mol.$
$⇒C\%_{MgCl_2}=\dfrac{0,3.95}{7,2+109,5-(0,3.2)}.100\%=24,55\%$
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